# If a map $f\colon S_g\to S_g$ induces the identity in $H_1(S_g,\mathbb{Z})$, then $f$ induces the identity in $H_1(S_g,\mathbb{Z}_2)$

all: This should not be too hard, but I am stuck. $S_g$ is the orientable, genus-g surface, and $H_1(S_g,\mathbb{Z})$ is the first homology with coeffs. in $\mathbb{Z}$. I am trying to avoid using the universal coefficient theorem, if possible.

I have been trying to find an answer using chains and mod2 reduction, i.e., using the fact that we can embed non-trivial $\mathbb{Z}/2$ $1$-chains $c$ into the group of $\mathbb{Z}$ $1$-chains, as just chains $c'$, with odd coefficients. Then $f$ will send the embedded $c'$ into a homologous ( in $H_1(S_g,\mathbb{Z})$ ) $\mathbb{Z}$ $1$-chain $c''$. But I don't know if mod 2 reduction will preserve homology, i.e., if $c'\sim c''$ , does it follow that $c'\pmod{2}\sim c''\pmod{2}$ ( i.e., reducing coefficients term-by-term)?

I know some properties of mod 2-reduction, but this is from $\mathrm{SL}(2,\mathbb{Z})$ to $\mathrm{SL}(2, \mathbb{Z}/n)$; kernel has finite index in $\mathrm{SL}(2,\mathbb{Z})$; it is finitely-generated by elements of finite order, but this does not seem to lead anywhere useful.

Maybe showing naturality of the map would help. Thanks.

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Sorry, I did not know how to do a tag to note that Z is the integers. – gary May 30 '11 at 2:40
I've added some LaTeX code; \mathbb{Z} produces the blackboard bold Z. You can see the code that generates the formulas by right clicking on a formula and selecting "Show source". – Arturo Magidin May 30 '11 at 2:45
Thanks, Arturo, I will add it next time, I appreciate it. – gary May 30 '11 at 2:53
A proper statement of the UCT mentions that the reduction map $H_*(X, \mathbb Z) \to H_*(X, \mathbb Z_2)$ is natural in maps of $X$. So the only thing you have to do to finish the theorem is prove the map is onto. – Ryan Budney May 30 '11 at 2:55
Ryan: Can I conclude this just at the chain level? I mean, if f_* maps c to c'(at chain level) with c~c'. Can I conclude that c1'~c1' , where c1:= c(mod2) andc1':=c'(mod2)? – gary Jun 12 '11 at 7:57